Problem: For a function $h$, we are given that $h(-4)=7$ and $h'(-4)=1$. What's the equation of the tangent line to the graph of $h$ at $x=-4$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $y+4=1(x-7)$ (Choice B) B $y-7=1(x+4)$ (Choice C) C $y-1=7(x+4)$ (Choice D) D $y+4=7(x-1)$
Solution: The derivative of a function at a point gives the slope of the line tangent to the function's graph at that point. Therefore, $h'(-4)$ gives the slope of the tangent line to the graph of $h$ where $x=-4$. We are given that $h'(-4)=1$, so the slope of the tangent line is $1$. Furthermore, we are given that $h(-4)=7$, which means the point of intersection of the tangent line and the graph is $(-4,7)$. To summarize, the tangent line has a slope of $1$ and it passes through the point $(-4,7)$. We can use the point-slope form of linear equations to find the tangent line equation: $\begin{aligned} y-y_1&=m(x-x_1) \\\\ y-7&=1(x-(-4)) \\\\ y-7&=1(x+4) \end{aligned}$ The equation is $y-7=1(x+4)$.